Lagrangian in the Path Integral

[YeaNah]

TL;DR Summary

Is the Lagrangian in the path integral for QFT actually classical? Since the inverse of the differential operator is taken to be the quantum propagator, shouldn't this mean the differential operator itself is a quantum equation?

Using free scalar field for simplicity.

Hi all, I have a question which is pretty simple, we have the path integral in QFT in the presence of a source term:

$$
Z[J] = \int \mathcal{D}\phi \, e^{i \int d^4x \left( \frac{1}{2} \phi(x) A \phi(x) + J(x) \phi(x) \right)}
$$

So far so good. Now this action is classical, right? Well when solve the integral in the presence of a classical source term we get:

$$ Z[J] = Z[0] e^{-\frac{i}{2} \int d^4x \, d^4y \, J(x) \mathcal{A}^{-1} J(y)} $$
where
$$\mathcal{A}^{-1} = D(x - y)$$

Here, ##\mathcal{A}^{-1} = D(x - y)## is commonly interpreted as a quantum propagator. But if this is so, then ##\mathcal{A}## must be the differential operator for the Klein-Gordon equation of a quantum scalar particle, not a classical Klein-Gordon field. Therefore, my question is, is the Lagrangian that goes into the path integral actually classical? Or is it in fact tied to single particle QM wavefunctions? Kind of like a sum over all possible histories the wavefunction can take?

Thanks.

$$\phi(x) = \int d^4y \, D(x - y) \mathcal{A} \phi(y).$$

 

[vanhees71]

There are some caveats! The "original path integral" is the Hamiltonian version. Often you can integrate out the canonical momenta (in your case the canonical field momenta, ##\partial \mathcal{L}/\partial{\dot{\phi}}##. Often the Hamiltonian is bilinear inthe potentials with field-independent coefficients. Then you get the Lagrangian version of the path integral with the Lagrangian given by its classical form. For details, see Sect. 4.5 and the beginning of Chpt. 6 in

https://itp.uni-frankfurt.de/~hees/publ/lect.pdf

 

[YeaNah]

Thanks for that. So my understanding is the original formulation has the quantum canonical momenta with the corresponding Hamiltonian, but we can get it in the form of a classical Lagrangian. Is this Lagrangain actually interpreted classically though ? I mean I know it’s equivalent to the classical Lagrangian of the free scalar field but since it’s the same Lagrangian that would give rise to the quantum Klein-Gordon equation through the Euler Lagrange equations can it be interpreted as a kind of quantum differential operator? It’s just the quantum propagator is making it hard for me to accept a purely classical interpretation of the differential operator.

 

[Demystifier]

Differential operators and differential equations are neither classical nor quantum. It is the physical interpretation that makes them classical or quantum. For example, if you say that a solution of the differential equation is a probability amplitude, then it's a quantum interpretation. But before you say anything about the interpretation, it doesn't make sense to even ask (let alone answer) whether it is classical or quantum. The same mathematical object can be classical in one interpretation and quantum in another.

 

[Demystifier]

Therefore, my question is, is the Lagrangian that goes into the path integral actually classical? Or is it in fact tied to single particle QM wavefunctions?

The ##\phi(x)## in the path integral is definitely not tied to a single particle wave function. In that sense it is "not quantum".

 

[YeaNah]

Differential operators and differential equations are neither classical nor quantum. It is the physical interpretation that makes them classical or quantum. For example, if you say that a solution of the differential equation is a probability amplitude, then it's a quantum interpretation. But before you say anything about the interpretation, it doesn't make sense to even ask (let alone answer) whether it is classical or quantum.

Okay thanks that makes sense. I guess my question would come down to the specific interpretation then of the differential operator in this context. Given that the inverse of the differential operator is interpreted as quantum, I guess it would make sense to interpret the differential operator itself in the path integral for QFT as quantum and it’s solutions as single particle wavefunctions? Perhaps the term ‘second quantisation’ isn’t so bad after all under this view. However it seems there are many who insist on a classical interpretation of the differential operator when it is first put into the path integral, which I find hard to reconcile with the inverse being the propagator. What are your thoughts?

 

[YeaNah]

The ##\phi(x)## in the path integral is definitely not tied to a single particle wave function. In that sense it is "not quantum".

Yes I think you’re right. Perhaps the answer is that the inverse of the differential operator is a term that ‘contributes’ the vacuum to vacuum amplitude, and isn’t strictly speaking an amplitude itself. After all it is describing virtual particles.

 

[Demystifier]

Okay thanks that makes sense. I guess my question would come down to the specific interpretation then of the differential operator in this context. Given that the inverse of the differential operator is interpreted as quantum, I guess it would make sense to interpret the differential operator itself in the path integral for QFT as quantum and it’s solutions as single particle wavefunctions? Perhaps the term ‘second quantisation’ isn’t so bad after all under this view. However it seems there are many who insist on a classical interpretation of the differential operator when it is first put into the path integral, which I find hard to reconcile with the inverse being the propagator. What are your thoughts?

Physicists often use a sloppy language. In the context of classical field theory, instead of talking about "propagator" it is more correct to talk about Green function. Mathematically they are the same, but they have a different physical interpretation. The propagator tacitly implies a quantum interpretation, while Green function tacitly implies a classical one. But physicists often don't care about the difference in their meaning.

 

[Demystifier]

Yes I think you’re right. Perhaps the answer is that the inverse of the differential operator is a term that ‘contributes’ the vacuum to vacuum amplitude, and isn’t strictly speaking an amplitude itself. After all it is describing virtual particles.

It looks as if you are generally confused about the relation between particles and fields. Perhaps my https://arxiv.org/abs/quant-ph/0609163 Secs. 7-9 may be helpful.

 

[YeaNah]

Physicists often use a sloppy language. In the context of classical field theory, instead of talking about "propagator" it is more correct to talk about Green function. Mathematically they are the same, but they have a different physical interpretation. The propagator tacitly implies a quantum interpretation, while Green function tacitly implies a classical one. But physicists often don't care about the difference in their meaning.

Combining my reply to both your messages into one. Your article looks very interesting, I haven’t ever viewed the first quantised harmonic oscillator in that way. So I guess the propagator is just a mathematical tool to get final amplitudes. As for the Green’s function being mathematically equivalent to the propagator, it’s just the interpretation that is different, I guess my question would be how do you personally interpret the differential operator then? If you had to choose, knowing that the inverse is a ‘quantum’ object

 

[vanhees71]

Physicists often use a sloppy language. In the context of classical field theory, instead of talking about "propagator" it is more correct to talk about Green function. Mathematically they are the same, but they have a different physical interpretation. The propagator tacitly implies a quantum interpretation, while Green function tacitly implies a classical one. But physicists often don't care about the difference in their meaning.

This is a bit subtle. Indeed the propagator used in perturbation theory, depicted by Feynman diagrams, is a Green function of the operator ##\Box+m^2##, which is, however not unique.

When we solve classical field equations, what we usually need is the retarded propagator, defined by
$$(\Box+m^2) G_{\text{ret}}(x)=-\delta^{(4)}(x), \quad G_{\text{ret}}(x) \propto \Theta(x^0).$$
For perturbation theory we need the propagator, defined as the time-ordered correlation function of the field operators wrt. the vacuum state,
$$\mathrm{i} G^{--}(x)=\langle \Omega |T \hat{\phi}(x) \hat{\phi}(0)|\Omega.$$
The difference in energy-momentum representation is apparently "tiny":
$$\tilde{G}_{\text{ret}}(p)=\frac{1}{(p^0+\mathrm{i} 0^+)^2-\vec{p}^2-m^2}$$
vs.
$$\tilde{G}^{--}(p)=\frac{1}{(p^0)^2-\vec{p}^2-m^2+\mathrm{i} 0^+}=\frac{1}{p^2-m^2+\mathrm{i} 0^+}.$$

 

[Demystifier]

When we solve classical field equations, what we usually need is the retarded propagator,

You mean retarded Green function, in classical field theory we should not call them "propagators". (Otherwise, I agree with everything you said above.)

 

[Demystifier]

I guess my question would be how do you personally interpret the differential operator then? If you had to choose, knowing that the inverse is a ‘quantum’ object

I interpret the differential operator as a purely mathematical object, devoid of any physical meaning.

 

[vanhees71]

Ok. For me Green's functions and propagators are just synonyms, but perhaps one should reserve "propagator" only for the time-ordered Green's function (or the Feynman Green's function, which are identical in vacuum QFT, but not in many-body QFT).